Question: $x^2-3xy+y^2=1$ Find the value of $\dfrac{dy}{dx}$ at the point $(1,0)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $1$ (Choice C) C $\dfrac{2}{3}$ (Choice D) D $-\dfrac{2}{3}$
We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $x^2-3xy+y^2=1$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} x^2-3xy+y^2&=1 \\\\ \dfrac{d}{dx}(x^2-3xy+y^2)&=\dfrac{d}{dx}(1) \\\\ \dfrac{d}{dx}(x^2)-3\dfrac{d}{dx}(xy)+\dfrac{d}{dx}(y^2)&=0 \\\\ 2x-3\left(1\cdot y+x\cdot\dfrac{dy}{dx}\right)+2y\cdot\dfrac{dy}{dx}&=0 \\\\ 2x-3y-3x\cdot\dfrac{dy}{dx}+2y\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 2x-3y-3x\cdot\dfrac{dy}{dx}+2y\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(2y-3x)&=3y-2x \\\\ \dfrac{dy}{dx}&=\dfrac{3y-2x}{2y-3x} \end{aligned}$ Now we can plug the point $(1,0)$ into the expression for $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{3y-2x}{2y-3x} \\\\ &=\dfrac{3(0)-2(1)}{2(0)-3(1)} \gray{x=1,\,\,y=0} \\\\ &=\dfrac{-2}{-3} \\\\ &=\dfrac{2}{3} \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at the point $(1,0)$ is $\dfrac{2}{3}$.